Find the frequency distribution for ordinary English, and use it to help you crack the code.

Dating Made Easier

Stage: 4 Challenge Level:

We have yet to receive a solution that uses
only Level 4 mathematics, (so a solution that doesn't use logs) and
so this problem has become Toughnut! See if you can crack it and
send us your solution!

We have now received some excellent
solutions for this Toughnut question. Patrick from Woodbridge
School solved the first question by making a list of each years
values:

Let x be the original value.
To add 10%, we multiply by 1.1, and we are trying to reach 2x.
Thus, we get
x
1.1x
1.21x
1.331x
1.4641x
1.61051x
1.771561x
1.9487171x
2.14358881x
So, after 8 years, the sum has doubled.

To reduce a value by 10%, we multiply by 0.9, and we must reach
0.5x. Thus, we get
x
0.9x
0.81x 11
0.729x
0.6561x
0.59049x
0.531441x
0.4782969x
so after 7 years the sum has halved in value.

Alexandra, Shannon, Gemma, Katie, Ruby and
Caitlin from Herts and Essex High School for Girls also worked out
the problem in this way correctly. Well done!

Iain from St. Gregory's Catholic High
and Daniel from King's College, Madrid used logarithms to solve the
problem:

If a sum increases by 10% each year this means that the value of
the previous year is multiplied by 1.1 (it is 110% of what it was
before). Therefore for an initial value "a" the progression will
be:
Year: 0
1
2
3
...
$a$
$a\times1.1$
$(a\times1.1)\times1.1$
$((a\times1.1)\times1.1)\times1.1)$
= $a$
$a\times(1.1)$
$a\times(1.1^2)$
$a\times(1.1^3)$
...

An object depreciates in value by 10% each year:
$0.9^m =
0.5$
(2)
$mlog0.9 = log0.5$
$m = \frac{log0.5}{log0.9} =
6.579$
(4.s.f.)
It will therefore be 7 years before the sum is effectively
halved.

Patrick from Woodbridge School went on to
explain why it takes different time for the value to double and
half:

Why aren't these two answers the
same?
The answers are not the same as, when 10% is added to a value, less
than 10% must be taken off as the 10% of the new number is larger.
This can be summed up as:
If $10$%$\times x + x = y$
then $y - 10$%$\times y < x$

Is there a rate, used for both
gain and depreciation, for which those two answers would actually
be the same?
For the above reason, there is no such rate as this is true for all
numbers and all percentages, except 0% and £0.

For a non-zero initial value, we cannot
find a non-zero rate for which the time taken for the value to
double and half would be the same. This is because for any
rate, the ratio of values between two successive years will be
different for gain and depreciation. To get the same time we really
need these ratios to be the same.

The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the
NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to
embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities
can be found here.